CHEMISTRY
A Molecular Approach
3rd Edition
Chapter 14
Chemical Equilibrium
Definition of EQUILIBRIUM
• When the rate of the forward reaction is
equal to the rate of the backward reaction.
• Rate of production of product is the same as
the rate of the reformation of the reactants.
The Concept of Equilibrium
• The double arrow implies the process is dynamic.
• Consider: A  B
Forward reaction: A  B Rate = kf[A]
Reverse reaction: B  A Rate = kr[B]
• At equilibrium kf[A] = kr[B].
The Concept of Equilibrium
• For an equilibrium we write A
• As the reaction progresses
B
– [A] decreases to a constant,
– [B] increases from zero to a constant.
– When [A] and [B] are constant, equilibrium is achieved.
• Alternatively:
– kf[A] decreases to a constant,
– kr[B] increases from zero to a constant.
– When kf[A] = kr[B] equilibrium is achieved.
The Concept of Equilibrium
The Equilibrium Constant
• For a general reaction
aA + bB
cC + dD
the equilibrium constant expression for everything in
solution is

C D
Kc 
a
b
A  B
c
d
where Keq is the equilibrium constant.
The Equilibrium Constant
• For a general reaction in the gas phase
aA + bB
cC + dD
the equilibrium constant expression is
c d
C D
a b
A B
P P
Kp 
P P
where Keq is the equilibrium constant.
Kp and Kc
• Kp = Kc(RT)Dn
or
Kc = Kp
RTDn
• Where n = # moles products minus
the # mole of reactants
2 TYPES OF EQUILIBRIUM
• 1. HOMOGENEOUS EQUILIBRIUM –
all components in the same state
• 2. HETEROGENEOUS EQUILIBRIUM
– components in different states
• DIFFERENCES IN K
EXPRESSION
Heterogeneous Equilibria
• In heterogeneous equilibria, pure solids and pure liquids
are not included in the equilibrium expression. Since the
concentration of pure liquids and solids is constant, they
will be given a value of 1.
• For example:
CaCO3(s)
CaO(s) + CO2(g)
HETEROGENEOUS EQUILIBRIA
• We ignore the concentrations of
pure liquids and pure solids in
equilibrium constant expressions.
• CaCO3 (s)  CaO (s) + CO2 (g)
• Keq = 1/[PCO2]
The Equilibrium Constant
• Kc is based on the molarities (Kp on partial pressure)
of reactants and products at equilibrium.
• We generally omit the units of the equilibrium
constant.
• Note that the equilibrium constant expression has
products over reactants.
• The same equilibrium is established no matter how
the reaction is begun.
The Magnitude of Equilibrium
Constants
• K = [products] ;
•
[reactants]
therefore
• If K >> 1; MORE products present at equil.
than reactants; Equilibrium lies to the right
• If K << 1; LESS products present at equil.
than reactants; Equilibrium lies to the left
The Equilibrium Constant
The Magnitude of Equilibrium
Constants
• If K << 1, then reactants dominate at equilibrium and the
equilibrium lies to the left.
The Equilibrium Constant
The Direction of the Chemical Equation
and Keq
• An equilibrium can be approached from any direction.
• Example:
N2O4(g)
2NO2(g)
• has
Kp 
2
NO2
P
PN 2O 4
 6.46
The Equilibrium Constant
The Direction of the Chemical Equation
and Keq
• In the reverse direction:
2NO2(g)
N2O4(g)
P
1
N 2O 4
K eq 
 0.155 
6.46
P2
NO 2
The Equilibrium Constant
Other Ways to Manipulate Chemical
Equations and Keq Values
• The reaction
2N2O4(g)
4NO2(g)
has
Kc 
4
NO 2
2
N 2O 4
P
P
which is the square of the equilibrium constant for
N2O4(g)
2NO2(g)
The Equilibrium Constant
Other Ways to Manipulate Chemical
Equations and Keq Values
• Equilibrium constant for the reverse direction is the
inverse of that for the forward direction.
• When a reaction is multiplied by a number, the
equilibrium constant is raised to that power.
• The equilibrium constant for a reaction which is the sum
of other reactions is the product of the equilibrium
constants for the individual reactions.
Sample Problems
• The equil. constant for the ff. rxn. is 0.013.
• 2NO (g) + Br2 (g) D 2NOBr(g)
• 1. Does the rxn favor the reactants or
products?
• 2. Calculate Keq for:
a. 2NOBr(g) D 2NO (g) + Br2 (g)
•
b. 4NO(g) + 2Br2 D 4NOBr(g)
Sample Problem
• Write the equilibrium expression for:
• NH3(aq) + H2O(l) D NH4+(aq) + OH-(aq)
Applications of
Equilibrium Constants
Predicting the Direction of Reaction
• We define Q, the reaction quotient, for a general reaction
aA + bB
cC + dD
as
Q
• Q = K only at equilibrium.
PCc PDd
PAa PBb
Applications of
Equilibrium Constants
Predicting the Direction of Reaction
• If Q > K then the reverse reaction must occur to reach
equilibrium (i.e., products are consumed, reactants are
formed, the numerator in the equilibrium constant
expression decreases and Q decreases until it equals K).
• If Q < K then the forward reaction must occur to reach
equilibrium.
• Please note that Q can only be calculated when initial
concentrations are given.
Calculating K
• 1. Balance the reaction.
• 2. Determine whether initial or equilibrium
concentrations are given.
• 3. If equil. concentrations are given, K can
be calculated directly using the equation
K= [product]n/[reactants]n.
• 4. If initial concentrations are given, use
the “ICE method”.
Problem
• N2(g) + 3H2(g) D 2NH3(g), Keq = 4.51 x 10-5
• Indicate whether the mixture is at equilibrium at
450 oC and in which direction the mixture must
shift to achieve equilibrium if:
• A.) 105 atm NH3, 35 atm N2, 495 atm H2
• B.) 26 atm NH3, 42 atm H2, 202 atm N2
• C.) 105 atm NH3, 55 atm H2, 5.0 atm N2
Calculating Equilibrium
Constants
• Proceed as follows:
– Tabulate initial and equilibrium concentrations (or partial
pressures) given.
– If an initial and equilibrium concentration is given for a species,
calculate the change in concentration.
– Use stoichiometry on the change in concentration line only to
calculate the changes in concentration of all species.
– Deduce the equilibrium concentrations of all species.
• Usually, the initial concentration of products is zero.
(This is not always the case.)
Calculating K
• H2(g) + F2(g) D 2HF(g)
• The equilibrium constant for the reaction
above is 115 at a certain temperature. In
a particular experiment, 3.00 mol of each
component was added to a 1.5 L flask.
Calculate the equilibrium concentration
for all species.
Problem
• At 327 oC, the equilibrium concentrations
are [CH3OH] = 0.15 M, [CO] = 0.24 M
and [H2] = 1.1 M for the reaction:
•
•
CH3OH (g) D CO (g) + 2H2 (g)
Calculate Kc at this temperature.
Problem
• Consider the following rxn. at a certain
temperature:
• Fe (s) + 3 O2 (g) D 2 Fe2O3 (s)
• An equil. mixture contains 1.0 mole Fe,
0.0010 mole O2 and 2.0 mole Fe2O3 all in a 2.0
L container. Calculate Kc for this reaction.
Problem
• A 1.00 L flask was filled with 2.00 mole
gaseous SO2 and 2.00 mole of NO2 and
heated After equil. was reached, it was
found that 1.30 mole gaseous NO was
present. Calculate the value of Kc for:
• SO2 (g) + NO2 (g) D SO3 (g) + NO (g)
Problem
• At a particular temperature, Kc = 4.0 x 10-7
for the reaction:
N2O4 (g) D 2NO2 (g)
In an expt. 1.0 mole of N2O4 is placed in a 1.00 -L
vessel. Calculate the concentrations of N2O4 and
NO2 when this reaction reaches equilibrium.
Use of Approximation
• Approximations can be used if the value
of K is so small that the terms (-x) or (+x)
will barely have an effect.
• Validity of approximation:
•
______[x]______
=
[initial concentration]
< 5%
< 5%
Le Châtelier’s Principle
• Le Châtelier’s Principle:
• If a system at equilibrium is disturbed,
the system will move in such a way as to
counteract the disturbance.
Le Châtelier’s Principle
Effects of Volume and Pressure Changes
Effect of Temperature Changes
• The equilibrium constant is temperature dependent.
• For an endothermic reaction, DH > 0 and heat can be
considered as a reactant.
• For an exothermic reaction, DH < 0 and heat can be
considered as a product.
Le Châtelier’s Principle
Effect of Temperature Changes
• Adding heat (i.e. heating the vessel) favors away from
the increase:
– if DH > 0, adding heat favors the forward reaction,
– if DH < 0, adding heat favors the reverse reaction.
• Removing heat (i.e. cooling the vessel), favors towards
the decrease:
– if DH > 0, cooling favors the reverse reaction,
– if DH < 0, cooling favors the forward reaction.
Le Châtelier’s Principle
The Effect of Catalysis
• A catalyst lowers the activation energy barrier for the
reaction.
• Therefore, a catalyst will decrease the time taken to reach
equilibrium.
• A catalyst does not effect the composition of the
equilibrium mixture.
• CH4 (g) + 2H2S (g)  CS2 (g) + 4H2 (g)
DH = -47 kJ
• What will be the effect on the pressure of H2 if:
•
•
•
•
•
A.
B.
C.
D.
E.
Volume of the container is increased?
CS2 gas is decreased?
A catalyst was added?
Heat was added?
Some CH4 were added?
Problem
• For the reaction at a particular temperature:
• 3 H2 (g) + N2 (g) D 2 NH3 (g)
• The equil. concentrations are [H2] = 5.0 M,
[N2] = 8.0 M and [NH3] = 4.0 M. What are
the initial concentrations of H2 and N2?
Sample Problem
• NH3(aq) + H2O(l) D NH4+(aq) + OH-(aq)
• Enough ammonia is dissolved in 5.00 L of
H2O at 25 oC to produce a 0.0124 M NH3
solution. Analysis of the equilibrium
mixture shows the OH- concentration to be
0.000464 M. Calculate Keq at 25 oC for the
reaction.
Sample Problem
• SO2Cl2(g) D
SO2(g) + Cl2(g)
• At equilibrium, the partial pressures of
SO2Cl2 = 3.31 atm, SO2 = 1.59 atm and
Cl2 = 0.98 atm.
• Calculate the value of the equilibrium
constant Kp.
Sample Exercise
• N2(g) + 3H2(g) D 2NH3(g), Kp = 1.45 x 10-5
• In an equilibrium mixture of the three gases at
500 oC, the partial pressure of H2 is 0.928 atm
and that of N2 is 0.432 atm. What is the partial
pressure of NH3 in the equilibrium mixture?
Quiz
• SO2Cl2(g) D SO2(g) + Cl2(g) Kc = 0.078
• In an equilibrium mixture of the 3 gases, the
concentration of SO2Cl2 = 0.108 M and SO2 =
0.052 M.
• A. Write the K expression.
• B. What is the partial pressure of Cl2 in the
equilibrium mixture?
Problem
• A mixture of 0.10 mole NO, 0.050 mole H2 and 0.10
mole of H2O is placed in a 1.0 L vessel at 300 K. The
following equilibrium is established.
• 2NO (g) + 2 H2 (g) D
N2 (g) + 2 H2O (g)
• At equilibrium [NO] = 0.062. Calculate the
equilibrium concentrations of H2, N2 and H2O.
• Calculate Kc.
Quiz
• 2SO2(g) + O2(g)D SO3(g) Keq = 0.345
• A. Write the K expression.
• B. In an equilibrium mixture , the partial
pressures of SO2 and O2 are 0.165 atm
and 0.755 atm, respectively. What is the
equilibrium partial pressure of SO3 in the
mixture?
Problem
• At 800 K, Kc = 3.1 x 10-5 for:
I2 (g) D
2I (g)
• If an equil. mixture in a 10.0 L vessel contains 2.67 x
10-2 grams of I (g), how many grams of I2 are in the
mixture?
• -----------------------------------------------------------------• 2SO2 (g) + O2 (g) D 2SO3 (g) Kp = 3.0 x 104
• In a 2.00-L vessel, the equil. mixture contains 1.57 g
SO3 and 0.125 g of O2. How many grams of SO2 are
in the vessel.
Problem
Br2(g) +Cl2(g) D 2BrCl(g) Kc = 7.0
At 400 K, if 0.30 mol of Br2 and 0.30 mol
of Cl2 are introduced into a 1.0-L
container, what will be the equilibrium
concentrations of Br2, Cl2 and BrCl?
Problem
• At 218 oC, Kc = 1.2 x 10-4 for the equilibrium
• NH4HS (s) D NH3 (g) + H2S (g)
• Calculate the equilibrium concentrations of
NH3 and H2S if a sample of solid NH4HS is
placed in a closed vessel and decomposes until
equilibrium is reached.
Problem
• At 25 oC, the reaction:
• CaCrO4 (s) D Ca 2+ (aq) + CrO4 2- (aq)
• has an equilirbium constant Kc = 7.1 x 10-4.
What are the equilibrium concentrations of
Ca2+ and CrO4 2- in a saturated solution of
CaCrO4?
Sample problem
• H2(g) + I2(g) D 2HI(g)
Keq = 51
• Predict in which direction the reaction will
proceed to reach equilibrium at 448 oC if we
start with 0.02 mol HI, 0.01 mol H2 and
0.03 mol I2 in a 2.00 L container.
Sample problem
• H2(g) + I2(g) D 2HI(g)
Keq = 50.5
• A 1.00 L flask is filled with 1.00 mol H2
and 2.00 mol I2 at 448 oC. What are the
partial pressures of H2, I2 and HI at
equilibrium?
Sample Problem
• 2NOCl(g) D
2NO(g) + Cl2(g) K = 1.6 x 10-5
• 1.0 mole NOCl is placed in a 2.0-L flask.
What are the equilibrium
concentrations?
Sample Problem
• N2(g) + 3H2(g) D 2NH3(g), Keq = 4.51 x 10-5
• Indicate whether the mixture is at equilibrium at
450 oC and in which direction the mixture must
shift to achieve equilibrium if:
• A.) 105 atm NH3, 35 atm N2, 495 atm H2
• B.) 26 atm NH3, 42 atm H2, 202 atm N2
• C.) 105 atm NH3, 55 atm H2, 5.0 atm N2
Le Châtelier’s Principle
• Le Châtelier’s Principle:
• If a system at equilibrium is disturbed,
the system will move in such a way as to
counteract the disturbance.
Le Châtelier’s Principle
•
•
•
•
Effects of Volume and Pressure
Changes
Le Châtelier’s Principle: if pressure is increased the
system will shift to counteract the increase.
That is, the system shifts to remove gases and
decrease pressure.
An increase in pressure favors the direction that has
fewer moles of gas.
In a reaction with the same number of product and
reactant moles of gas, pressure has no effect.
Le Châtelier’s Principle
Effects of Volume and Pressure Changes
Effect of Temperature Changes
• The equilibrium constant is temperature dependent.
• For an endothermic reaction, DH > 0 and heat can be
considered as a reactant.
• For an exothermic reaction, DH < 0 and heat can be
considered as a product.
Le Châtelier’s Principle
Effect of Temperature Changes
• Adding heat (i.e. heating the vessel) favors away from
the increase:
– if DH > 0, adding heat favors the forward reaction,
– if DH < 0, adding heat favors the reverse reaction.
• Removing heat (i.e. cooling the vessel), favors towards
the decrease:
– if DH > 0, cooling favors the reverse reaction,
– if DH < 0, cooling favors the forward reaction.
Le Châtelier’s Principle
The Effect of Catalysis
• A catalyst lowers the activation energy barrier for the
reaction.
• Therefore, a catalyst will decrease the time taken to reach
equilibrium.
• A catalyst does not effect the composition of the
equilibrium mixture.
• CH4 (g) + 2H2S (g)  CS2 (g) + 4H2 (g)
DH = -47 kJ
• What will be the effect on the pressure of H2 if:
•
•
•
•
•
A.
B.
C.
D.
E.
Volume of the container is increased?
CS2 gas is decreased?
A catalyst was added?
Heat was added?
Some CH4 were added?